$f(x)=x^2$ $f'(x)=$
Answer: $f$ is of the form $x^n$ and therefore we can apply the power rule: $\dfrac{d}{dx}[x^n]=n\cdot x^{n-1}$ $\begin{aligned} f'(x)&=\dfrac{d}{dx}[x^{{2}}] \\\\ &={2}x^{{2}-1} \\\\ &=2x \end{aligned}$ In conclusion, $f'(x)=2x$